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Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.

 Apr 26, 2022
 #1
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That question has been solved here: https://web2.0calc.com/questions/algebra-question_67

 Apr 26, 2022
 #2
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(a^2)(b^3)=(20)^18.

 

20^18 =2^36 * 5^18

a =(2^18)^2 =262,144^2

b =(5^6)^3  =15,625^3

 

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 Apr 26, 2022
 #3
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20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered pairs= 2,701

 

20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered triplets= 4,104,919

 Apr 26, 2022
 #4
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a=262,144

b=15,625

 

LCM[15,625, 262,144)=11,790,000

11790000 = 2^4 * 3^2 * 5^4 * 131

 

Number of ordered pairs=[ (4*2+1) + (2*2+1) + (4*2+1) + (2*1+1)]==9 * 5 * 9 *3==1,215

 Apr 26, 2022
 #5
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Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.

 

\( 20^{18}=2^{36}*5^{18}\)

 

 

\(5^0*5^{18}\\ 5^6*5^{12}\\ \\~\\ 2^0*2^{36}\\ 2^6*2^{30}\\ 2^{12}*2^{24}\\ 2^{18}*2^{18}\\ \)

 

5^0 can be paired with 7 of them  (I mean different powers of 2)

5^6 can be paired with 7 of them

 

So I just get 14  but order counts, and a and b are all interchangeable so  that is 28

 

Here they are

If you can spot any that I have missed then please let me know.

(I am not guarenteeing that this answer is right)

 Apr 27, 2022
edited by Melody  Apr 27, 2022

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